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CHRISTINE BREINER: Welcome
back to recitation.

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In this video, I'd
like us to work

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on the following problem that
has to do with tangent planes

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and approximations.

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So we know that a rectangle--
we'll say a rectangle

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has sides x and y, and
we know that to find

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the area of the rectangle
then, we just take x times y,

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and I would like us to
approximate the area for x

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equal to 2.1 and y equal to 2.8.

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And obviously, with
this type of equation,

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it's not hard to
just compute this,

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but I'd like us to use the
tangent plane approximation

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to determine the
value, and then we'll

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compare it to the actual
value, just to give us

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an idea of how we can
use the tangent plane

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to approximate things.

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And obviously, I'd like to
do this near x equal 2 and y

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equal 3.

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So when you're doing your
tangent plane approximation,

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do the approximation at
x equal 2 and y equal 3.

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And then when you're
done with that,

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I would like you to
answer this question.

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So near x equal 2 and y equal
3, which has the greater effect?

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A change in x or an
equal change in y?

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So if I change by
x in the same value

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as I change by y, which will
have the greater impact?

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So why don't you pause the
tape, work on the problems,

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and when you're ready to see
how I do them, bring it back up,

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and I'll come back and show you.

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OK, welcome back.

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So we're going to work on this
problem, and as I mentioned,

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the first thing we want to
do is do a tangent plane

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approximation near x
equal 2 and y equal 3.

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So let me write down
what we're going

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to need in order to do that.

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First, we'll remind ourselves
that in this case, what

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I'm going to
approximate is the area

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function for a rectangle,
which is A of x,

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y is equal to x times y.

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So that's the function I'm
going to be approximating.

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And now the actual
approximation,

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the tangent plane approximation,
has the following form.

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So we know A of x,
y is approximately--

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well, it's going to be the
area evaluated at the point I'm

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interested in, which
we said was 2 comma 3,

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right, plus the x-derivative
of area evaluated at 2

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comma 3, times the change in x.

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And the change in x is where x
is now versus where we started,

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which is at x equal 2.

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And then the same
thing in terms of y.

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So you do the y-derivative
evaluated at 2 comma 3,

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and then you do the change in y.

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And since we started at y
equals 3, the change in y

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is y minus 3, OK?

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So now we obviously need to
fill in three quantities here.

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The three things we need
to evaluate: area evaluated

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at (2, 3), its x-derivative
evaluated at (2, 3),

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and its y-derivative
evaluated at (2, 3).

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So let me just point
out what we have.

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Area evaluated at (2, 3),
well, that's just 2 times 3,

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which we can do, so that's 6.

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That's pretty easy.

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Now A sub x, so the
derivative of the area

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function with respect to x.

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The derivative of the area
function with respect to x

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is just y.

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y in that case we
treat as a constant.

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We take this derivative.

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We just get y back.

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And so, A sub x
evaluated at (2, 3)

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is going to be y,
which-- y here is 3,

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so we get A sub x is
equal to 3 at (2, 3).

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In a similar vein, we can
immediately look and see

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that A sub y evaluated at
(2, 3) is going to be 2.

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And the reason for
that, of course,

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is if we look back
here, the derivative

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of A with respect to y is x,
so evaluating it at (2, 3)

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gives us 2.

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OK.

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So now we just have
to fill everything in.

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So my tangent
plane approximation

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now says I get 6 plus 3 times x
minus 2 plus 2 times y minus 3.

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And what I wanted was the
area at 2.1 comma 2.8.

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So if I fill in those
values for x and y--

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so 2.1 is x and 2.8 is
y-- if I fill those in,

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I get that this is
equal to-- well, I'll

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keep writing
approximately just to be

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safe-- is 6 plus 3 times--
well, 2.1 minus 2 gives me a 0.1

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there and 2 times 2.8 minus 3
gives me a negative 0.2 there,

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so I get a negative 0.4,
I get a positive 0.3,

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so together this
is a negative 0.1.

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6 minus 0.1 gives me 5.9.

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So the area based
on the approximation

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is 5.9 square units.

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And, in actuality, if
you multiply it out,

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I think you get
something like 5.88,

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so the approximation
is very good, right?

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We're close to
(2, 3) and that is

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one of the reasons we can
know it's going to be very,

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it should be very good.

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It should be pretty good, OK?

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Now, I just have to answer the
second part of the question.

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So let me remind us what
the second part was.

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It was near x equal
2 and y equal 3,

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which has a greater effect?

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A change in x or an
equal change in y?

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And to do that, it's really
easiest to come back and look

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at maybe this line.

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I'll underline this
line right here.

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Actually, I'll box it, OK?

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This value will represent
the change in x.

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So we started at 2,
and we go somewhere,

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and that'll represent
the change in x.

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This value represents
the change in y.

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So if we look at which has a
greater impact near (2, 3),

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if my change in x and my
change in y are equal,

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then obviously this term
has a bigger impact,

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because there's a 3.

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The coefficient here is 3 and
the coefficient here is 2.

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And so the point is, changes
in x will have a greater

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effect than changes in y.

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And what this corresponds
to pictorially

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is if we make a slice
where we keep y constant

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and we look at the curve in
the xz-plane, that corresponds

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to the fact that the derivative
in the xz-plane of the curve

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there is more significant--
the curve there

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has a steeper derivative-- than
if I kept the x-value fixed

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and I looked in the yz-plane
and I looked at the curve there.

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So the sort of
one-dimensional picture

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is that the derivative of
a curve in the x-direction,

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keeping y fixed, is
steeper than the derivative

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of the curve in the
y-direction, keeping x fixed.

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So maybe-- hopefully,
that wasn't more confusing

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than it should have been.

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From here, you can
see it right away,

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but that's sort of the
picture of it as well.

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So I think that's
where I'll stop.

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